Kerala-engg-paper2-solution-2015.pdf

SOLUTIONS & ANSWERS FOR KERALA ENGINEERING ENTRANCE EXAMINATION-2015 – PAPER II* VERSION – B1 [MATHEMATICS] 1.

Ans:

Sol:

0<x≤

9.

7 3

7 − 3x ≥ 0 ⇒ x ≤ 0<x≤

Ans:

−1 + i

Sol:

i10 + i11 + i12 + i13 = 0 1 1 1 1 = 11 + 12 + 13 + 14 i i i i ∴ Remaining expression = −1 + i

7 and x ≥ 1 3

7 . 3

10. Ans: 2.

Ans:

f(2n + 1) = −1 for all n n f(1) = f(2) = 1 and f(2 ) = 1, (i.e) ∴ f(4) = f(8) = f(16) = 1 f(2n) = −1 when n is odd ∴ sum = 5 − 20 = −15.

4.

5.

Ans:

X=Y

Sol:

X=Y

Ans:

(A ∪ B)\A ∩B

Sol:

(A ∪ B)\A ∩B

Ans:

Sol:

6.

7.

8.

12. Ans:

1   3 , 1  

Sol:

−1 ≤ cos3x ≤ 1 −1 ≤ −cos3x ≤ 1 1 ≤ 2 − cos3x ≤ 3 1 1 1≥ ≥ . 2 − cos 3 x 3 28

Sol:

n(C) = 40, n(F) = 16, n(H) = 11 n(C ∩ F) = 8, n(C ∩ H) = 6 n(F ∩ H) = 2, n(C ∩ F ∩ H) = 1 n(C ∪ F ∪ H) = 67 − 16 + 1 = 52 n(C′ ∩ F′ ∩ H′) = 80 − 52 = 28

Ans:

3

Sol:

1 − 3y = −26, y = 9 |y| = 3.

Ans:

x = 1 − 2y

2

Sol:

14. Ans: Sol:

15. Ans:

x2 + y 2 = −y + 1

⇒ x + y = y − 2y + 1 2 ⇒ x = 1 − 2y. 2

13. Ans:

2

2

2

2

2

f(z) = 1 + z + z 2 2 Re(f(z)) = 1 + x − y + x 2

Sol:

Ans:

Sol:

2

Re f (z ) = x − y + x + 1= 0 2 2 x − y + x + 1 = 0.

11. Ans: 3.

2

−15 Sol:

Sol:

x −y +x+1=0

Sol:

2

π 2 2

z = 2i 2 z i 1+i i Z e − = 2i e − = 2i e = 2e[i] π Arg = . 2 p+q+1=0 4p − 4q = 4(q − p) 2 2 p −q +p−q=0 ⇒p+q+1=0 2

2

6 |x − 3| = 1 x − 3 = ±1 x = 2, 4 2

2x + 9x + 45 = 0 3 9 ⇒ 3(α + β) = − 2 2 5 45 αβ = ⇒ 9αβ = 2 2 9 45 x2 + x + =0 2 2 2 2x + 9x + 45 = 0.

α+β= −

−1 and 1 11 y=

x

x − 5x + 9 2 x y − (5y + 1)x + 9y = 0 2

for real x, ∆ ≥ 0 2 2 i.e (5y + 1) − 36y ≥ 0 (5y + 1 − 66) (5y + 1 + 6y) ≥ 0 (−y + 1) (11y + 1) ≥ 0 (y − 1) (11y + 1) ≤ 0  −1  ⇒ y ∈  , 1  11  16. Ans: Sol:

17. Ans: Sol:

22. Ans:

Sol:

ab + bc + ca 3x − 2(a + b + c)x + ab + bc + ca = 0 2 4(a + b + c) = 12(ab + bc + ca) 2 (a + b + c) = 3(ab + bc + ca) 2 2 2 a + b + c = ab + bc + ca. 2

2x + 2x − 1 = 0. 2

1 1+ 3

=

1− 3 1 3 =− + −2 2 2

1 3 1 Sum = −1, product = − = − 4 4 2 1 2 x +x− =0 2 2 2x + 2x − 1 = 0.

23. Ans: Sol:

24. Ans: Sol:

25. Ans:

1   50 1 − 10   2  50 25 + + ..... area S10 4 4 1 a = 25, r = 2 1   25 1 − 10   2  = 50 1 − 1  .   1  210  1− 2

25 +

2 2

16a c

4

4

2 2

(a + c) = (2b) = 16a c 10

11 24(x + 1) = 3 (x − 2)(x − 1) x = 10. 3 10

18. Ans: Sol:

19. Ans: Sol:

20. Ans:

Sol:

1 3 4

21. Ans: Sol:

= 1 + 10C1 .

ar = 1458, ar = 54 1 1 3 . ⇒r = ⇒r= 27 3 900 S1 + S2 + …… + S9 = 45 S10 + ………+ S19 = 55 S20 + ……… + S29 = 65 S30 + ……… + S39 = 75 S40 + ………. + S49 = 55 ………………. S (99) = 135 S = 5 (180) = 900.

1 1 . x + 10C2 . . x 2 + 2 4

1 3 10 1 4 . x + C4 . . x + .... 8 16 1 By inspection, 10C3 . is the highest 8 coefficient . ∴ power of the x is 3. 10

7

26. Ans: Sol:

27. Ans: Sol:

C3 .

5880, No correct option given. No. of ways = 7 × 7 × 6 × 5 × 4 = 5880 13 (m + 4) (m + 5) = 22(m − 1) m = 7, 6 m1 + m2 = 13

3

28. Ans: Sol:

x  1 +  2 

a11 + a12 + a13 = 141 a21 + s22 + s23 = 261 x + 161 = 141 2 ⇒ x = 21 ⇒ 3(a + d) = 21 a+d=7 3a + 63d = 261 ⇒ a = 3. 2

Sol:

29. Ans:

1024 211 = 210 = 1024 . 2 120 n

Sol:

n

n

r

=

Cr −1

16

a ar = 32, r = 2 4 4 ⇒ a = 4 ⇒a = 4 ⇒r=2 14 16 a15 = 4 × 2 = 2 .

Cr Cr

n

Cr −1

15

n

3

n − (r − 1) r

= n + 1− r 15

∑ r n Cr −r 1 = ∑ (16 − r ) 1

C

1

= 16 x 15 −

1 . 15 × 16 2

= 240 − 120 = 120. 30. Ans: Sol:

31. Ans: Sol:

37. Ans:

−∆

Sol:

∆ = (36 − 30) −2(24 − 15) + 3(12 − 9) = −3 ∆′ = 4(−6) −8(−9) + 15(−3) = 3 = −∆.

C1 → C1 + C2 + C3 9+x 3 5 ⇒ 9+x 2+ x 5

9+x

Sol:

=0

1 3 ⇒ (9 + x) 0 x − 1 0

5 0

=0

x −1

0

39. Ans: Sol:

⇒ (9 + x) (x − 1) = 0 ⇒ x = −9 or x = 1, 1 2

32. Ans: Sol:

33. Ans:

−6

40. Ans:

−3(12 − 4) + 2b(−2) = 0 −24 −4b = 0 4b = −24 b = −6

Sol:

41. Ans:

−3 ,3 2 |A| = x − 2x(x − 1) = −9 2 ⇒ x − 2x + 2x = −9 2 ⇒ 2x − 3x − 9 = 0

3 ± 9 + 72 3 ± 9 = 4 4 −3 x = 3 or . 2

42. Ans: Sol:

Sol:

0

43. Ans:

C1 → C1 + C2 + C3 0 cos2 36°

cot 135°

⇒ 0 cot 35° sin2 37° = 0 0 cos2 25° cos2 65° 35. Ans: Sol:

36. Ans: Sol:

~(p ∨ q) ∨ (~p ∧ q) = (~p ∧ ~q) ∨ (~p ∧ q) = ~p ∧ (~q ∨ q) = ~p ∧ T = ~p (~p ∧ ~q) ∨ (~p ∧ ~r) ~[p ∨ (q ∧ r)] = ~p ∧ ~(q ∧ r) = ~p ∧ (~q ∨ ~r) = (~p ∧ ~q) ∨ (~p ∧ ~r) ~(P ∨ Q) ~(P ∨ Q) 4 5

ABA

Sol:

1 2 sinθ = = θ 1 1 + tan2 1+ 2 4 1 4 = = . 5 5 4 2×

4 (x − 5) − (y − 5) = 4sec θ − 4tan θ = 4 2

44. Ans: Sol:

(−2, ∞)

2

2

2

4 tan15 + tan75 =2−

(ABA)′ = A′B′A′ = ABA

|x − 3| < 2x + 9 ⇒ −(2x + 9) < (x − 3) < 2x + 9 −2x −9 < x − 3 ⇒ −3x < 6 ⇒ x > −2 ----(1) x − 3 < 2x + 9 ⇒ −x < 12 ⇒ x > −12 --(2) From (1) and (2) we get x > −2

θ 2

( )

x=

34. Ans:

~p

2 tan

Sol: Sol:

A = xy, P = 2(x + y) 2 Since (x + y) ≥ 4xy 2

x+4

3

2

P ⇒   ≥ 4A 2 2 ⇒ P ≥ 16A.

38. Ans:

1, 1, −9

P ≥ 16A

3 +2+ 3 = 4 .

π f(x) = cos4x + tan3x 2π π Period of cos4x = = 4 2 π Period of tan3x = 3

π π ∴ period of f(x) = LCM of  ,  3 2 = π.

45. Ans:

n +1 n −1

Sol:

46. Ans:

Sol:

sin(θ + φ) = n sin(θ − φ) sin(θ + φ) n ⇒ = sin(θ − φ ) 1 sin(θ + φ) + sin(θ − φ) n + 1 Nr + Dr ⇒ = Nr − Dr sin(θ + φ) − sin(θ − φ) n − 1 2 sin θ cos φ n + 1 ⇒ = 2 cos θ sin φ n − 1 tan θ n + 1 ⇒ = . tan φ n − 1

sin x − sin x + 3sin x − 3sin x = 0 4 4 2 (sin x + 3) (sin x − sin x) = 0 4 2 sin x = sin x sinx = 0 3 solutions 2 sin x = 1 2 solutions 5 solutions. 8

50. Ans:

6

4

2

(0, 4)

Sol:

C(h, k)

 16  tan−1   13  2 tan

(4, 2) •

1 1 + tan−1 3 4

−1

1   2.  3  + tan−1 1  = tan 1  4  1−  9  3  1 = tan−1  + tan−1  4 4 −1

A(0, 0)

h+8 k +0 ,   = (4, 2 ) 2   2

 3 1  +   4 4  = tan 3 1   1− .  4 4 

⇒ h = 0, k = 4 ∴ C(0, 4).

−1

51. Ans:

 16  = tan−1  .  13 

47. Ans:

Sol:

2π 7 π  π  2π ⇒  − sin−1 x  +  − sin−1 y  = 2  2  7 ⇒ sin

48. Ans:

Sol:

49. Ans: Sol:

x + sin

−1

2π 5π y = π− = . 7 7

52. Ans: Sol:

 1   − 1, 2  

:

1

G

C(5, 5)

A, B, C are collinear points x1 − 3 = x2 − x1 y1 − 4 = y2 − y1 y − 4 y 2 − y1 ⇒ 1 = x1 − 3 x 2 − x1 ∴ slope of AB = slope of BC ∴ A, B, C are collinear.

cos−1 x > sin−1 x π ⇒ cos−1 x > − cos−1 x 2 π ⇒ 2 cos−1 x > 2 π cos−1 x > 4 1 x< and x > −1. 2

53. Ans:

Sol:

a

2

p1 =

6

sin x + cos x = 1 8 2 3 sin x + (1 − sin x) = 1

−a 2

sec θ + cos ec 2θ

= a sin θ cos θ =

− a cos 2θ

a sin 2θ 2

= a cos 2θ cos2 θ + sin2 θ 2 2 2 2 2 2 ⇒ (2p1) + (p2) = a ⇒ 4p1 + p2 = a . p2 =

5 8

2

 10 + 2 10 + 2  G ,  = (4, 4 ) 3   3

cos−1 x + cos−1 y =

−1

(4, 4) • O(2, 2)

Sol:

5π 7

B(8, 0)

54. Ans:

17 3

Sol:

Solve : 2x + y = 2 x + 2y = 2 ---------------2 2 x= , y= 3 3 2

d=

55. Ans: Sol:

56. Ans: Sol:

2

2 17  2  . 1 −  +  2 −  = 3 3 3    

60. Ans:

The required line is 4x − y = k which passes through (5, 0) ⇒ 20 − 0 = k ⇒ k = 20 4x − y = 20 x y − =1 5 20 x y + =1 5 − 20 ∴ (0, −20).

2

=

2

61. Ans:

58. Ans: Sol:

64. Ans:

g2 + f 2 − c = 9 + 1 + 28 = 38

 2 +1  ,  2 

2 + 1  2 

Distance between the direction = 2 2 .3

Choice (D)

Sol:

centre : (h, k) = (−g, −f) = (3, −1) ∴ x = h + r cosθ, y = k + r sinθ

3 2

2

63. Ans:

2g = −6 ⇒ g = −3 2f = 2 ⇒ f = 1 c = −28

2

x + y = a(x + y)

=

x = 3 + 38 cos θ, y = −1 + 38 sin θ

r=

59. Ans:

Sol:

= 4 + 16

= 2 5.

2

2

Equation of the circle passing through (0, 0) (a, 0) (0, a) 2 2 x + y + 2gx + 2fy + c = 0 ––––– (1) passing through (0, 0) ⇒ c = 0 passing through (a, 0) ⇒ 2g = –a passing through (0, a) ⇒ 2f = –a ∴ Equation of the circle 2 2 x + y – ax – ay = 0 2 2 ⇒ x + y = a(x + y) Choice (C)

62. Ans:

20

2

⇒ (x − 3) + y = 16

1 k k k2 × × = = 10 2 5 1 10

(8 − 6 )2 + (− 2 − 2)2

(10cosθ + 10sinθ)

 5(x − 3 )   y  ⇒   +  =1  20   4 

k 5 k y intercept = 1

solve x + y = 6 x + 2y = 4 --------------⇒ y = −2, x = 8 ⇒ (8, −2) ∴ centre : (8, −2)

3

20 cos θ + 15 20 sin θ + 0 , y= 5 5 ⇒ 5x − 15 = 20cosθ, 5y = 20sinθ

x intercept =

57. Ans: 2 5

:

• P(x, y)

x=

Sol:

±10

r=

2 • (5, 0)

⇒ k = ±10.

Sol:

a circle

Sol:

(0, −20)

Area =

 2 +1 2 + 1   , is the only option satisfy  2 2   the equation of the circle 2 2 (x − 1) + (y − 1) = 1.

Sol:

Sol:

0, –4 y = mx + c be a tangent to 2 2 x = 4ay is c = –am 1 ⇒ k = − k 2 ⇒ k =0, –4 4 x2 3y2 + =1 9 8 2b2 16 8 = ⇒ b2 = a 9 3 Equation of ellips a = 3,

x 2 y2 + = 1⇒ 8 9 3

a e

=3 2

x2 3y2 + =1 9 8

65. Ans: Sol:

AC = AB + AD = 3i +6j + 2k D

40 a=5 b=4 4 Max {cp} + 5min {cp) = 4 × 5 + 5 × 4 = 40

66. Ans: Sol:

Sol:

A AC ∴ P= AC

(0, –1) 2

=

2

y – 4x – 2y – 3 ⇒ (y –1) = 4 (x + 1) 2 shift the origin to (–1, 1) ⇒ y = 4x focs, (1, 0), referring the original axis, focus is (0, 1) ∴ Equation of Latusection is x = 0 2 ∴ y – 2y – 3 = 0 ⇒ y = –1 or 3 ∴ and points on (0, –1), (0, 3)

9 5 7

71. Ans:

Sol:

Sol:

⊥or distance from a2 or r = a1 + λ b is

b = 2i − 2 j + 2k

d

2−2+2 12

=

i

2

(a2 − a1)× b = 3

3. 2 3

j k 0 0 = −18 j + 9k

2 3 6

1 = 3 –1  1  θ = cos   3

405 9 5 = 7 7

∴d=

72. Ans:

α = ± 1, β = 1

Sol:

Sol:

2i+3j+6k (4, 2, 2)

a. b a b 3

68. Ans:

(1, 2, 2)

b =7

a = i+ j+k

=

(a2 − a1) × b b

–1  1  cos   3

cosθ =

B

1 (3i + 6 j + 2k ) 7

d=

67. Ans:

C

1 1 1 4 3 4 = 0⇒ β = 1 1 α β

0

a+b

2

= 4 + 4 + λ2 = 8 + λ2

a−b

2

= 8 + λ2

∴ a +b − a −b = 0

1+ α + β = 3 ⇒ α + β = 2 ⇒ α = 1 2

2

2

2

73. Ans:

∴α=±1

69. Ans: Sol:

2

14

Sol:

PQ = OQ − OP ⇒

0 Q=

∴QA = a −

PQ + OP = OQ ∴ PQ + OP = OQ =

70. Ans:

1 (3i + 6 j + 2k ) 7

a+b+c 3

=

14

a+b+c 3

2a − b + c 3

∴ QA + Q B + Q C = 0

74. Ans: Sol:

π 2 Given lines are x y 2 x y z = = and = = 1 1 −1 1 −1 −1 2 3 6 4

1 1 1 1 . + . × −1 + 1× = 0 2 6 3 4 π θ= 2

75. Ans:

1 – cos α = 2

sin α = 2

11 3

82. Ans: Sol:

a b Projection = b

Sol:

=

76. Ans:

22 11 = 6 3

2 2   sin−1   3   

=

3 4

P A ∩B

)

P(A ∪ B) = P(A) + P(B) – P (A ∩ B) ⇒ P(A) + P(B) = P(A ∪ B) + P (A ∩ B) ∴ 1 + P(A ∩ B) – P(B) – P(A) = 1+ P(A ∩ B) – (P(A ∪ B) + P(A ∩ B) C = 1 – P(A ∪ B) = P (A ∪ B) = P A ∩B

(

)

83. Ans: 1 and 7 Sol:

Let the unknown date be a and b 1 Variance = x 2 − x 2 = 5 .2 n 1 ⇒ 4 + 16 + 36 + a2 + b2 − 16 = 5.2 5 2 2 ⇒ a + b = 50 Also 2 + 4+ 6 + a + b = 20 ⇒ a + b = 8 It is clear that the two observations are 1 and 7

∑

Ιf θ is the required angle then sin θ = cosin if angle between i − j + k and 2i − j + k

Sol:

(

3 4

(

2 + 1+ 1

2 2 = 3 3 6

2 2   θ = sin−1   3   

)

84. Ans: 24 77. Ans: –2 Sol: Sol:

cos2α + cos2β + 1 – 2sin γ = 1 cos2α + cos2β + cos2γ = –1 ∴ value = –2 ∴ choice (C)

sum of first n terms =

Midpoint joining (3, 2, 6) and (5, 4, 8) (i) (4, 3, 7) satisfying x + y +z = 14

Sol:

79. Ans: (1, 3, 5) Sol:

Foot of the ⊥ lie on the line. Satisfying only A(1, 3, 5) r

80. Ans: 3 Sol:

Sol:

45 4 There are a total of 18 observations a 1 × 10 + (a2 × 8 ) ∴ combined mean = , 18 where a1 and a2 denote the means But a1 = 15 ⇒ 40 150 + 8a2 45 = ⇒ a2 = 3 18 4

(

)

86. Ans: 26

The required plane is (2 + 3λ)x + (−8 −5λ)y + (4 + 4λ)z + −(p + 10λ) = 0 Compare with the plane x + 3y + 0z + 13 = 0 comparing coefficients λ = −1 Then we get p = 3.

Sol:

87. Ans: 81. Ans:

n (296 + (n − 1) (− 2)) 2

= n × 125 (given) ∴ 298 – 2n = 250 ⇒ n = 24

85. Ans:

78. Ans: x + y +z = 14 Sol:

a = 148 ; d = –2

2

3 4 cos 60 + cos 45 + cos α = 1 1 1 + + cos2 α = 1 4 2 2

2

2

Sol:

2

f(256) = f(16 ) = f(16) + 6 2 f(16) = f(4 ) = f(4) + 6 2 f(4) = f(2 ) = f(2) + 6 But f(2) = 8 ∴ f(4) = 14; f(16) = 20 and f(256) = 26 1 2 10 9 8 7 6 5 4 3 1 . . . . . . . . 9 10 7 8 5 6 3 4 2 = 1 2

88. Ans: 3 Sol:

∴

The domain is not explicitly given; it is R’ − {0, ± 1} and the function is continuous on its domain

97. Ans:

89. Ans: 1 Sol:

Sol:

90. Ans:

Sol:

 1    sin y  1  lim it = lim  −  y →α   1  y  y     =1−0=1 2 9 limit = lim

x →α

=−

2x 2 = 3(3 x − 2) 9

=

Sol:

f’(x) = g(x + 1) f’’(x) = g’(x + 1) = h(x) f’’(2x) = h (2x)

 1 − cos 2 x   f(x) = sin−1   2 sin x  1 = sin− (sin x) f’(x) = ±1 |f’(x) = 1

Sol:

2

y = 100 + 45 = 145 Yy’ = 0 ⇒ y’ = 0 (Q y ≠ 0)

Sol: x →0

= −7

Sol:

+

2 x3

1 + 2 at x = 1 4

7 4

[3 sin (10x + 11) −7] is maximum 2 When sin (10x + 11) is minimum 2 ∴ minimum value of 3sin (10x + 11) = 0 2 ∴ Maximum value = (−7) = 49. 2

If A sq. units in the area measure when the radius is r units, 2 their A = πr dA dr = 2πr ∴ dt dt dA dr But = 3c dt dt ∴ 2πr = 3c ∴ c = 4π, when r = 6

dy = 6x − 5 dx = 6 – 5 = 1 at (1, 4)

101. Ans: 8

x Sol:

1+ x2 y = sec(tan− x)

227 − 5 222 = = 37 11 − 5 6 6x = 48 ⇒ x = 8

6x – 11 =

1

= sec sec 1 + x 2

102. Ans: k ≥ 1

y = 1 + x2 dy x = dx 1+ x2

Sol:

f’(x) = cosx – k ∴ f decreases if cosx ≤ k ∴ for decreasing k ≥ 1

Choice (A)

103. Ans: –3 96. Ans: 1 Sol:

2

100. Ans: 1

The limit = lim (3x − 7)

95. Ans:

−2 (1 + x )3

99. Ans: 4π

94. Ans: −7 Sol:

=

98. Ans: 49

93. Ans: 0 Sol:

1 1 + 1+ x +1 x dy 1 1 = − dx (1 + x )2 x 2

y = 1−

dx 2

92. Ans: 1 Sol:

7 4

d2 y

91. Ans: h (2x) Sol:

dx =1 dy

Sol:

Ιf t = tan θ,

sin x = sin2θ, tan y = tan 2θ, |θ| < π ∴ x = 2θ, y = 2θ

4

dy m = + 2nx + 1 dx x m ∴ + 2n 2 + 1 = 0 2 m + 2n + 1 = 0

m + 8m + 2 = 0 ∴ 6n + 1 = 0

109. Ans:

1 n=− 6n + 1 = 0  6 ∴ ⇒ 3m + 2 = 0 m = − 2 3 −4 5 2m + 10n = − 3 3 = –3

Sol:

 xn  1 tan−1  + C  a  na  

xn −1dx

1

du

∫ x2n + a2 = n ∫ a2 + u2 ; if a = x

n

=

1 1 u . tan−1 + C n a a

=

 xn  1 tan−1  + C  a  na  

104. Ans: 3

–1

Sol:

110. Ans: log|x| + 2tan x + C

acceleration time t = 12t – 12 ∴ acceleration is zero when t = 1 2 velocity at time t = 6t – 12t + a ∴ –3 = 6 – 12 + a ⇒a=3

Sol:

∫

cot ( x e x )

=

∫

if u = x e

x

Sol:

= log |sec u| + C x = log |sec (x e )| + C

106. Ans:

Sol:

∫

dx x log x

∫

1 = 3

1 + x3

∫ ∫

2

=

=

(u − 1)

du

∫

=−

2 1 + x 3 ( x 3 − 2) + C 9 x5



1 log log x 2 + C 2

111. Ans: =

dx cot u

2

–1

x

(1 + x )e x

1

∫ x(x2 + 1) dx = ∫  x + x2 + 1 dx = log|x| + 2tan x + C

105. Ans: log |sec (x e )| + C Sol:

(x + 1)

2

=

3

if u = 1 + x

du (if u = log x) 2u 1 log u + C' 2

1 log log x + C' 2

1 log log x 2 + C 2

u

 1   u−  du  u  

1 1 −1 1  1 −1 1    tan   − tan    9 4 4 5  5 

112. Ans: =

=

1 3

=

1  1  2 32 2  u − 2u  + C 3 3 

1

Sol:

2 1 + x3 (1 + x3 − 3 ) + C 9 2 = 1 + x 3 ( x 3 − 2) + C 9

=

∫

dx ( x 2 + 16 ) ( x 2 + 25 )

0 1

1 = 9

∫

1   1 − 2  2  dx  x + 16 x + 25 

0

107. Ans: log |2 e − 5| + C 2x

Sol:

∫

4 ex 2 e x − 5 e− x

1

dx = =

∫ ∫

4 ex 2 e2 x − 5

1 1 x 1 x . tan−1 − tan−1  9 4 4 5 5 0 dx

du 2x if u = 2 e − 5 u

113. Ans: 0

= log |u| + C 2x = log |2 ex − 5| + C

1

Sol:

108. Ans:

x2 + 2 x + log x + C 2

∫

∫

x(1 − x )(1 + x ) dx

−1

2

Sol:

1 1 1 1 1 tan −1 − tan−1  9  4 4 5 5

=

 1  1   x+  dx =  x + 2 + dx  x x    1 = x 2 + 2 x + log | x | +C 2

∫

The integrand is an odd function & the range is (−1, 1) l ∴ =0

∫

114. Ans: 4 Sol:

=−

3

Where the curves meet, 2 X = 2x ⇒ x \ 0 or 2

=

π 4 2

+

1 2

1  π 1 −  4 2

2

∫

∴ required area-measure

( x 2 − 2x )dx )

117. Ans: yy"+(y')2 + 1 = 0

0

Sol:

8 = −4 3 4

2

2

x +y =1 ∴2x + 2yy’ = 0 1 + y'2 + yy' ' = 0

3

118. Ans: y − 3xy = C 3

115. Ans:

π 2

Sol:

π

Sol:

π

∫

sin2 x

−π

1+ 7

x

dx =

0

∫

sin2 x dx

∫

1+ 7x

1+ 7

−π π

=

−π

1+ 7x π

= 2.

x

−

∫

−π

sin2 x 1 + 7− x

dx i.e, x.y =

116. Ans:

sin x dx

119. Ans: 3 and 2 Sol:

Sol:

1 π . 2 2

120. Ans: y = Cxe

∫ 0

Sol:

π

2

2 x sin x dx = 3

2

4

∫

The equation is

 2 d2 y  dy 2   d3 y 2   +   =   dx 2  dx    dx 3   

sin2 x dx

1  π 1 −  2  4 π

y 2dy + A

3

0

= 2.

∫

3xy = y − C

2

2

∫

The equation: dx Ιf + x = y2 dy

x sin x dx

x

dy y = y+ dx x dy dx = + dx y x log |y| = log |x| + x + A x y = Cx e

0 π

= [x( − cos x ) − ( − sin x )]0 4

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